Question: The equation of a circle $C$ is $x^2+y^2-4x-16y+64 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2-16y) = -64$ $(x^2-4x+4) + (y^2-16y+64) = -64 + 4 + 64$ $(x-2)^{2} + (y-8)^{2} = 4 = 2^2$ Thus, $(h, k) = (2, 8)$ and $r = 2$.